- #1

2Dark4u

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## Homework Statement

The Setup is depicted in the Attachment. Essentially there is a 5kg Mass on an incline (45). The incline measures 0.4m from the mass to the corner were it turns 90 degrees and measures 0.3m to were there is a spring (k = 20N/m. This spring attaches to the Mass. This system is in equilibrium with the spring stretched out as it is.

The question is what is the length of the Spring not stretched. As in if the mass was un attached and the spring was at rest?

## Homework Equations

Only relevant equation i can think of is F = k(λx) Basic Hookes law

## The Attempt at a Solution

So first thing I did

5* 9.8 = 49N

Then 49cos(45) to get Fx of mass = 34.65N this should equal the X component of the force of the spring.

Second the spring should measure 0.5 since it makes a triangle right triangle with 0.4 and 0.3 this also means the angle its at in relation to the x component of force is 36.87

so F of the spring should be Fx = F cos(36.87)

34.65N / cos( 36.87) = 43.31N

then F = k λx and F/k = λx

43.31/20 = 2.16m

Now if the spring stretch measures 0.5m I don't see how λx can equal 2.16m

That would mean that the spring has a negative measurement when unstretched and that doesn't make any sense.

So help me out, were did I go wrong.